Chapter 2, Problem 5P

(a)

To determine

To evaluate: The limit of the function $\underset{x\to 0}{\lim }\frac{〚x〛}{x}$.

Answer to Problem 5P

The limit of the function does not exist.

Explanation of Solution

Definition used:

Greatest integer function:$〚x〛=$ the largest integer that is less than or equal to x.

Calculation:

Obtain the limit of the function $\underset{x\to 0}{\lim }\frac{〚x〛}{x}$.

Let the function $f\left(x\right)=\frac{〚x〛}{x}$.

For $0<x<1$,

By the definition of greatest integer function, $〚x〛=0$.

Thus, $\frac{〚x〛}{x}=0$. (1)

For $-1<x<0$,

By the definition of greatest integer function, $〚x〛=-1$.

Thus, $\frac{〚x〛}{x}=\frac{-1}{x}$. (2)

The left hand limit of the function as x approaches 0 is computed as follows,

$\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{〚x〛}{x}\\ =\underset{x\to 0^{-}}{\lim }\left(-\frac{1}{x}\right)\left[\text{by equation (2)}\right]\\ =\frac{1}{0}\\ =\infty \end{array}$

The right hand limit of the function as x approaches 0 is computed as follows,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{〚x〛}{x}\\ =\underset{x\to 0^{+}}{\lim }\left(0\right)\left[\text{by equation (1)}\right]\\ =0\end{array}$

Since the left and right hand limits are not equal.

Thus, the limit of the function $\underset{x\to 0}{\lim }\frac{〚x〛}{x}$ does not exist.

(b)

To determine

To evaluate: The limit of the function $\underset{x\to 0}{\lim }x〚\frac{1}{x}〛$.

Answer to Problem 5P

The limit of the function is 1.

Explanation of Solution

Result used: Squeeze theorem

“Suppose that $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ for all x in some open interval containing c except possibly at c itself, if $\underset{x\to c}{\lim }g\left(x\right)=L=\underset{x\to c}{\lim }h\left(x\right)$ then $\underset{x\to c}{\lim }f\left(x\right)=L$”.

Calculation:

Obtain the limit of the function $\underset{x\to 0}{\lim }x〚\frac{1}{x}〛$.

Let the function $f\left(x\right)=x〚\frac{1}{x}〛$.

For $x>0$ such that,

$\begin{array}{c}\frac{1}{x}-1\le 〚\frac{1}{x}〛\le \frac{1}{x}\\ x\left(\frac{1}{x}-1\right)\le x〚\frac{1}{x}〛\le x\left(\frac{1}{x}\right)\\ x\left(\frac{1-x}{x}\right)\le x〚\frac{1}{x}〛\le x\left(\frac{1}{x}\right)\\ 1-x\le x〚\frac{1}{x}〛\le 1\end{array}$

Take the right hand limit of the above inequality as x approaches 0,

$\underset{x\to 0^{+}}{\lim }\left(1-x\right)\le \underset{x\to 0^{+}}{\lim }x〚\frac{1}{x}〛\le \underset{x\to 0^{+}}{\lim }1$

Here, $\underset{x\to 0^{+}}{\lim }\left(1-x\right)=1\text{ and }\underset{x\to 0^{+}}{\lim }\left(1\right)=1$.

Therefore, by the Squeeze theorem, $\underset{x\to 0^{+}}{\lim }x〚\frac{1}{x}〛=1$.

For $x<0$ such that

$\begin{array}{c}\frac{1}{x}-1\ge 〚\frac{1}{x}〛\ge \frac{1}{x}\\ x\left(\frac{1}{x}-1\right)\ge x〚\frac{1}{x}〛\ge x\left(\frac{1}{x}\right)\\ x\left(\frac{1-x}{x}\right)\ge x〚\frac{1}{x}〛\ge x\left(\frac{1}{x}\right)\\ 1-x\ge x〚\frac{1}{x}〛\ge 1\end{array}$

Take the left hand limit of the above inequality as x approaches 0,

$\underset{x\to 0^{-}}{\lim }\left(1-x\right)\ge \underset{x\to 0^{-}}{\lim }x〚\frac{1}{x}〛\ge \underset{x\to 0^{-}}{\lim }1$

Here, $\underset{x\to 0^{-}}{\lim }\left(1-x\right)=1\text{ and }\underset{x\to 0^{-}}{\lim }\left(1\right)=1$.

Therefore, by the Squeeze theorem, $\underset{x\to 0^{-}}{\lim }x〚\frac{1}{x}〛=1$.

Since the left and right hand limits are equal, the limit exists and equal to 1. That is, $\underset{x\to 0}{\lim }x〚\frac{1}{x}〛=1$.

Thus, the limit of the function is 1.